Solve for x 7^(3-5x)=(1/49)^(2x+9)

$7^{3 - 5 x} = {(\frac{1}{49})}^{2 x + 9}$

Step 1

Apply the product rule to $\frac{1}{49}$ .

$7^{3 - 5 x} = \frac{1^{2 x + 9}}{49^{2 x + 9}}$

Step 2

One to any power is one.

$7^{3 - 5 x} = \frac{1}{49^{2 x + 9}}$

Step 3

Move $49^{2 x + 9}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$7^{3 - 5 x} = 49^{- (2 x + 9)}$

Step 4

Create equivalent expressions in the equation that all have equal bases.

$7^{3 - 5 x} = 7^{2 (- (2 x + 9))}$

Step 5

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$3 - 5 x = 2 (- (2 x + 9))$

Step 6

Solve for $x$ .

Simplify $2 (- (2 x + 9))$ .

Rewrite.

$3 - 5 x = 0 + 0 + 2 (- (2 x + 9))$

Simplify by adding zeros.

$3 - 5 x = 2 (- (2 x + 9))$

Apply the distributive property.

$3 - 5 x = 2 (- (2 x) - 1 \cdot 9)$

Multiply.

Multiply $2$ by $- 1$ .

$3 - 5 x = 2 (- 2 x - 1 \cdot 9)$

Multiply $- 1$ by $9$ .

$3 - 5 x = 2 (- 2 x - 9)$

Apply the distributive property.

$3 - 5 x = 2 (- 2 x) + 2 \cdot - 9$

Multiply.

Multiply $- 2$ by $2$ .

$3 - 5 x = - 4 x + 2 \cdot - 9$

Multiply $2$ by $- 9$ .

$3 - 5 x = - 4 x - 18$

Move all terms containing $x$ to the left side of the equation.

Add $4 x$ to both sides of the equation.

$3 - 5 x + 4 x = - 18$

Add $- 5 x$ and $4 x$ .

$3 - x = - 18$

Move all terms not containing $x$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$- x = - 18 - 3$

Subtract $3$ from $- 18$ .

$- x = - 21$

Divide each term in $- x = - 21$ by $- 1$ and simplify.

Divide each term in $- x = - 21$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 21}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 21}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 21}{- 1}$

Simplify the right side.

Divide $- 21$ by $- 1$ .

$x = 21$

Solve for x 7^(3-5x)=(1/49)^(2x+9)

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